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How do you find the x and y intercepts for #y= 2x^2+3x - 5#?

1 Answer

Jun 17, 2016

#x# intercept #x=-5/2 or x=1#
#y# intercept #y=-5#

Explanation:

To find #y# intercept lying on y_axis so set #x=0# in the given equation :

Thus ,
#y=2(0)^2+3(0)-5#
Hence,#y=-5#

To find #x# intercept we have to set #y=0# and solve the equation to find #x#

#2x^2+3x-5=0#
#=(2x+5)(x-1)#

#2x+5=0#
#2x=-5#
#x=-5/2#

or #x-1=0#
#x=1#

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